Problem: A rectangle has a perimeter of 64 inches and each side has an integer length. How many non-congruent rectangles meet these criteria?
Solution: Call the height $h$ and the width $w$. We want to find the number of solutions to $2(w+h)=64$ or $w+h=32$. The solutions to this are  \[
\{(1,31),(2,30),\ldots,(16,16),\ldots,(31,1)\}.
\] There are 31 solutions to this, but we are double-counting all the rectangles for which $w \neq h$. There are 30 of these, so the total number of rectangles is $\frac{30}{2}+1=\boxed{16}$ rectangles.